### jeqcho's blog

By jeqcho, history, 3 years ago, Sometimes, you are asked to calculate the combination or permutation modulo a number, for example $^nC_k \mod p$. Here I want to write about a complete method to solve such problems with a good time complexity because it took me a lot of googling and asking to finally have the complete approach. I hope this blog can help other users and save their time when they solve combinatorics problem in Codeforces.

#### Example Problem

Find the value of $^nC_k$, ($1 \leq n,k \leq 10^6$). As this number can be rather large, print the answer modulo $p$. ($p = 1000000007 = 10^9 + 7$)

#### Combination (binomial coefficients)

$^nC_k$ means how many ways you can choose $k$ items from an array of $n$ items, also denoted as $\binom{n}{k}$. This is also known as binomial coefficients. The formula for combination is

$^nC_k = \frac{n!}{k!(n-k)!}$

Sometimes, the denominator $k!(n-k)!$ is very large, but we can't modulo it since modulo operations can't be done independently on the denominator. $\frac{n!}{k!(n-k)!} \mod p \neq \frac{n! \mod p}{k!(n-k)! \mod p}$. Now I will introduce the modular multiplicative inverse to solve this problem.

#### Modular multiplicative inverse

The modular multiplicative inverse $x$ of $a$ modulo $p$ is defined as

$a \cdot x \equiv 1 \pmod p$

Here, I will replace $x$ with $\text{inv}(a)$, so we have

$a \cdot \text{inv}(a) \equiv 1 \pmod p$

Getting back to the formula for combination, we can rearrange so that

$^nC_k = n! \cdot \frac{1}{k!} \cdot \frac{1}{(n-k)!}$

Here, we can use $\text{inv}(a)$ as follows

$^nC_k \equiv n! \cdot \text{inv}(k!) \cdot \text{inv}((n-k)!) \pmod p$

Now we can distribute the modulo to each of the terms by the distributive properties of modulo

$^nC_k \mod p = n! \mod p \cdot \text{inv}(k!) \mod p \cdot \text{inv}((n-k)!) \mod p$

Now I will discuss on how to calculate $\text{inv}(a)$

#### Fermat's Little Theorem

You can easily remember this theorem. Let $a$ be an integer and $p$ be a prime number,

$a^p \equiv a \pmod p$

It is helpful to know that the $p$ in the problem ($10^9 + 7$) is indeed a prime number! We can rearrange the equation to get

$a^{p-1} \equiv 1 \pmod p$

Looking back at our equation for $\text{inv}(a)$, both equations equate to 1, so we can equate them as

$a \cdot \text{inv}(a) \equiv a^{p-1} \pmod p$

We can rearrange the equation to get

$\text{inv}(a) \equiv a^{p-2} \pmod p$

We now have a direct formula for $\text{inv}(a)$. However, we cannot use the pow() function to calculate $a^{p-2}$ because $a$ and $p$ is a large number (Remember $1 \leq n,k \leq 10^6$) ($p = 10^9 + 7$). Fortunately, we can solve this using modular exponentiation.

#### Modular Exponentiation

To prevent integer overflow, we can carry out modulo operations during the evaluation of our new power function. But instead of using a while loop to calculate $a^{p-2}$ in $O(p)$, we can use a special trick called exponentiation by squaring. Note that if $b$ is an even number

$a^b = (a^2)^{b/2}$

Every time we calculate $a^2$, we reduce the exponent by a factor of 2. We can do this repeatedly until the exponent becomes zero where we stop the loop. This will give us a time complexity of $O(\log p)$ to calculate $a^{p-2}$ because we halve the exponent in each step. For the case when $b$ is odd, we can use the property

$a^b = a^{b-1} \cdot a$

We then store the trailing $a$ into a variable. Then $b-1$ is even and we can proceed as previously stated. We can repeatedly apply these two equations to calculate $a^{p-2}$. Here I will show you the implementation of this modified powmod() function to include modulo operations. ll is defined as long long

ll powmod(ll a, ll b, ll p){
a %= p;
if (a == 0) return 0;
ll product = 1;
while(b > 0){
if (b&1){    // you can also use b % 2 == 1
product *= a;
product %= p;
--b;
}
a *= a;
a %= p;
b /= 2;    // you can also use b >> 1
}
return product;
}


Then we can finally implement the $\text{inv}(a)$ function simply as

ll inv(ll a, ll p){
return powmod(a, p-2, p);
}


Then, finally, we can implement $^nC_k$ as

ll nCk(ll n, ll k, ll p){
return ((fact[n] * inv(fact[k], p) % p) * inv(fact[n-k], p)) % p;
}


We used the dp-approach for factorial where the factorial from 1 to n is pre-computed and stored in an array fact[].

#### Time complexity

• Pre-computation of factorial: $O(n)$
• Calculation of $^nC_k$, which is dominated by modular exponentiation powmod: $O(\log p)$
• Total: $O(n + \log p)$

#### Problems for you

Please comment below if you know similar problems.  Comments (4)
 » 3 years ago, # | ← Rev. 2 →   Good tutorial! It's worth mentioning that you can find modular inverse using Extended Euclidean Algorithm in $O(\log{p})$, too.Also, if $n$ and $k$ are small, you can calculate binomial coefficients with DP in $O(nk)$ without modular inverse.And here are some problems: